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Let $R$ be a commutative ring with 1, we define

$$N(R):=\{ a\in R \mid \exists k\in \mathbb{N}:a^k=0\}$$

and

$$U(R):=\{ a\in R \mid a\mbox{ is invertible} \}.$$

Could anyone help me prove that if $a\in N(R) \Rightarrow 1+a\in U(R)$?

I've been trying to construst a $b$ such that $ab=1$ rather than doing it by contradiction, as I don't see how you could go about doing that.

Bill Dubuque
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Freeman
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    If $a\in N(R)$, then we can find $k\in\mathbb N^*$ such that $a^k=0$. What about $\sum_{j=0}^{k-1}(-a)^j$? – Davide Giraudo Sep 28 '11 at 10:44
  • @LHS You may want to change your notation: I would not use $a$ to describe $U(R)$ and then say if $a \in N(R)$ ... –  Sep 28 '11 at 11:36
  • @DBLim Do not know him sorry, but I appear to have 5 mutual friends with him, Wadham i'm guessing? – Freeman Sep 28 '11 at 13:13
  • I know Adam from a few years ago when I met him at Warwick. Then he was doing his A-levels. I am no longer friends with him on the internet, but a last check reveals we have 16 mutual friends. –  Sep 28 '11 at 13:50
  • @DBLim Small world ;) I'll keep a look out for him in my lectures! – Freeman Sep 28 '11 at 14:20

5 Answers5

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This really is a problem in disguise: How did you derive the formula for the sum of the geometric series in year (something) at school??

$(a + 1)(a^{k-1} - a^{k-2} + \ldots 1) = 1-(-a)^n$

but as $a^n = 0$, we have (it does not matter whether $n$ is even or odd) that $(a+1)$ is invertible. Prove the following analogous problem, it may strengthen your understanding:

Let $A$ be a square matrix. If $A^2 = 0$, show that $I - A$ is invertible.

If $A^3 = 0$, show that $I - A$ is invertible.

Hence in general show that if $A^n = 0$ for some positive integer $n$, then $I -A$ is invertible.

J. W. Tanner
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  • Ah this is excellent, I don't think I would have spotted that! Need to get my game back I think ;) Thanks! – Freeman Sep 28 '11 at 13:09
  • @LHS Sometimes maths gives the illusion that everything is so complicated; rings, nilradicals, commutative rings, etc and the simple ideas get lost under a heap of terminology.... –  Sep 28 '11 at 13:48
  • Indeed, just an extra thing, how'd you think this lets you deduce that is $u\in U(R)$ and $a\in N(R) \Rightarrow u+a\in U(R)$? What i'm currently working on! – Freeman Sep 28 '11 at 14:19
  • As U(R) forms a group with multiplication i've been trying to multiply $a+u$ by things to get $u(a+1)=au+u \in U(R)$ or as a factor.. just use that fact in some way – Freeman Sep 28 '11 at 14:23
  • @LHS When $n$ is odd, $a^n + b^n = (a+b)(a^{n-1} + \ldots b^{n-1})$. There might be some alternating signs in there. –  Sep 28 '11 at 21:50
  • Ah I think I got it another way, $(a+u)u^{-1}=au^{-1}+1$. As $(au^{-1})^k=a^k(u^{-1})^k=0(u^{-1})^k \Rightarrow au^{-1}\in N(R) \Rightarrow au^{-1}+1\in U(R)$ – Freeman Sep 29 '11 at 07:23
  • @LHS By the way I just noticed that the question you asked is the first exercise in Atiyah Macdonald. –  Oct 05 '11 at 12:57
  • Is the "geometric series" supposed to be $(a+1)(1-a+a^2-a^3+...a^{\color{red}n-1})=1 \color{red}+ (-a)^n$? The overall argument still works – J. W. Tanner Oct 27 '24 at 18:50
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Note the following:

$(1+a)(1-a) = 1-a^2$.

$(1-a^2)(1+a^2) = 1-a^4$

$(1-a^4)(1+a^4) = 1-a^8$

...

Thus, continuing in this way, we may find some $b_n$ such that $(1+a)b_n = 1-a^{2^n}$

For large enough $n$, this will give us $(1+a)b_n = 1$.

the L
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  • can you explain the downvote? is this incorrect? – the L Sep 28 '11 at 11:25
  • No, this is correct, just not obviously how so. $(1+a)$ is a factor of $(1-a^{2^n})$ for $n\geq 1$. If we pick the smallest $n$ such that $a^{2^n} = 0$, taht will yield your result. – Arthur Sep 28 '11 at 11:54
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$N(R)$ is at least in some texts referred to as the nilradical of $R$. It is contained in all prime ideals (in fact, it is the intersection of all prime ideals, Atiyah, MacDonald prop. 1.8), so taking a nilpotent element $a$, since it is contained in all maximal ideals, $a+1$ is not in any maximal ideal. Then the ideal generated by $a+1$ must necessarily be the whole ring, which means that it specifically generates 1 at some point.

J. W. Tanner
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Arthur
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Let $a\in N(R)$ and $k$ such that $a^k=0$. We have \begin{align*}(1+a)\left(\sum_{j=0}^{k-1}(-a)^j \right)&=\sum_{j=0}^{k-1}(-a)^j+\sum_{j=0}^{k-1}-(-a)^{j+1}\\\ & =\sum_{j=0}^{k-1}(-a)^j-\sum_{j=1}^k(-a)^j\\\ &=1-(-a)^k=1+(-1)^ka^k=1, \end{align*} and we have $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)=1$ by the same computation (it's true even if the ring is not commutative), hence $1+a$ is invertible, and its inverse is $\left(\sum_{j=0}^{k-1}(-a)^j\right)(1+a)$.

J. W. Tanner
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Davide Giraudo
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    This seems a very complicated way of deriving a formula learnt at school... –  Sep 28 '11 at 11:34
  • @DBLim: Indeed, yours does explain the thought process needed to come to the conclusion, I guess this is another way of expressing the same idea.

    Thanks though, this is helpful!

     – Freeman Sep 28 '11 at 13:10
  • @DavideGiraudo Cette question n'a rien à voir, vous étudiez à quelle université maintenant?

    Vous avez dit que l'anglais n'est pas votre première langue, le même que le franćais n'est pas le mien. On peut se discuter en anglais/franćais pour que...

     –  Sep 28 '11 at 13:55
  • @DavideGiraudo Merci de me dire. –  Sep 28 '11 at 21:50
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Below is a (nonconstructive) more conceptual structural view (vs. the common constructive proof using geometric series - which also can be presented more conceptually via simpler multiples).

Hint $ $ A nilpotent $\rm\,n\,$ lies in every prime ideal $\rm\,P,\,$ by $\rm\, n^k = 0\in P\ \Rightarrow\ n\in P.\,$ In particular, this implies $\rm\,n\,$ lies in every maximal ideal. Hence $\rm\,n\!+\!1\,$ is a unit, since it lies in no maximal ideal $\rm\,M\,$ (else $\rm\,n\!+\!1,\,n\in M\, \Rightarrow\, (n\!+\!1)-n = 1\in M),\,$ i.e. elements coprime to every prime are units.

You may recognize this idea in modern proofs of Euclid's theorem that there are infinitely many primes. Namely, if there are only finitely many primes then their product $\rm\,n\,$ is divisible by every prime, so $\rm\,n\!+\!1\,$ is coprime to all primes, so it must be the unit $1,\,$ contradiction.

Remark $ $ You'll meet related results later when you study the structure theory of rings. There the intersection of all maximal ideals of a ring $\rm\,R\,$ is known as the Jacobson radical $\rm\,Jac(R).\,$ The ideals $\rm\,J\,$ with $\rm\,1+J \subset U(R)= $ units of $\rm R,\,$ are precisely those ideals contained in $\rm\,Jac(R).\,$ Indeed, we have the following theorem, excerpted from my post on the fewunit ring theoretic generalization of Euclid's proof of infinitely many primes.

Theorem $\ $ TFAE in ring $\rm\,R\,$ with units $\rm\,U,\,$ ideal $\rm\,J,\,$ and Jacobson radical $\rm\,Jac(R).$

$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\,J\,$ lies in every max ideal $\rm\,M\,$ of $\rm\,R.$

$\rm(2)\quad 1+J \subseteq U,\quad\ $ i.e. $\rm\, 1 + j\,$ is a unit for every $\rm\, j \in J.$

$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad $ i.e. proper ideals survive in $\rm\,R/J.$

$\rm(4)\quad M\,$ max $\rm\,\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\,R/J.$

Proof $\, $ (sketch) $\ $ With $\rm\,i \in I,\ j \in J,\,$ and max ideal $\rm\,M,$

$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\,$ unit.

$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\,$ unit $\rm\,\Rightarrow I = 1.$

$\rm(3\Rightarrow 4)\ \,$ Let $\rm\,I = M\,$ max.

$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\,$ by $\rm\,M\,$ max.

Bill Dubuque
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